Ta có : 1.98 + 2.97 + 3.96 + ...+ 98.1 = 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + .....+ ( 1 + 2 + 3 + ...+ 97 + 98 ) = \(\frac{1.2}{2}\)+ \(\frac{2.3}{2}\)+  \(\frac{3.4}{2}\)+ ...+ \(\frac{98.99}{2}\)= \(\frac{1}{2}\)( 1 . 2 + 2 . 3 + 3 . 4 +...+ 98 . 99).

Vậy A = \(\frac{1}{2}\)

Nè bạn giải cụ thể chi tiết cho mình đk k thì mình mới k cho đk

A=3/1.2+3/2.3+3/3.4+3/4.5+...+3/2021.2022

A=3(1/1.2+1/2.3+1/3.4+1/4.5+...+1/2021.2022)

A=3(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/2021-1/2022)

A=3[1/1+(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+...+(1/2021-1/2021)-1/2022]

A=3[1/1+0+0+0+...+0-1/2022

A=3(1/1-1/2022)

A=3(2022/2022-1/2022)

A=3.2021/2022

A=2021/674

Bn Tham Khảo:

https://hoc247.net/hoi-dap/toan-6/tinh-tong-s-3-1-2-3-2-3-3-3-4-3-4-5-3-2015-2016-faq188428.html

           S=1.2+2.3+3.4+4.5+...+98.99+99.100

suy ra :3S=1.2.3+2.3.3+3.4.3+4.5.3+...+98.99.3+99.100.3

            3S=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+98.99.(100-97)+99.100.(101-98)

           3S=1.2.3.0+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+98.99.100-97.98.99+99.100.101-98.99.100

           3S=99.100.101

Suy ra :S=99.100.10:3=333300

vậy S=333300

ko bit

=333300

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+5.6.(7-4)+6.7.(8-5)+7.8.(9-6)+8.9.(10-7)+9.10.(11-8)

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+5.6.7-4.5.6+6.7.8-5.6.7+7.8.9-6.7.8+8.9.10-7.8.9+9.10.11-8.9.10

=9.10.11

=> A=9.10.11:3

=3.10.11

=330

3A= 3.(1.2 + 2.3 + 3.4 + 4.5 + 5.6 + 6.7 + 7.8 + 8.9 + 9.10)
= 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + 4.5.(6 - 3) + 5.6.(7 - 4) + 6.7.(8 - 5) + 7.8.(9 - 6) + 8.9.(10 - 7) + 9.10.(11 - 8)
= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 - … + 8.9.10 - 8.9.10 + 9.10.11
= 9.10.11 = 990.
A= 990/3 = 330

\(A=1.2+2.3+3.4+4.5+5.6+6.7+7.8+8.9+9.10\)

\(A=2.\left(1+3\right)+4.\left(3+5\right)+6.\left(5+7\right)+8.\left(7+9\right)+9.10\)

\(A=8+32+72+128+90\)

\(A=330\)

`S = 1.2 + 2.3 + 3.4 + 4.5 + ... + 99.100.`

`3S =  1.2.3 + 2.3.(4-1) + 3.4.(5-4) + 4.5.(6-3) + ... + 99.100.(101-98)`

`3S =  1.2.3 + 2.3.4-1.2.3 + 3.4.5-4.5.6 + 4.5.6-3.4.5 + ... + 99.100.101-98.99.100`

`3S =  99.100.101`

`S = 33.100.101`

`S = 333300`

3S=1.2(3-0)+2.3(4-1)+.....+99.100(101-98)

=1.2.3-0.1.2+2.3.4-1.2.3+4.5.6-2.3.4+....+99.100.101-98-99-100

=99.100.101

S=33.100.101

=333300

S = 1.2 + 2.3 + 3.4 + ... + 99.100

3S = 1.2.3 + 2.3.3 + 3.4.3 + .. +99.100.3

3S = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)

3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

3S = 99.100.101

3S = 999 900

  S = 333 300

Vậy S = 333 300

=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2011.2012.3

=> 3S = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 2011.2012.( 2013 - 2010 )

=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + .... + 2011.2012.2013 - 2010.2011.2012

=> 3S = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 2010.2011.2012 - 2010.2011.2012 ) + 2011.2012.2013

=> 3S = 2011.2012.2013

=> S = ( 2011.2012.2013 ) : 3

3S=1.2.3+2.3.(4-1)+...............+2011.2012.(2013-2010)

3S=1.2.3+2.3.4-1.2.3+...............+2011.2012.2013-2010.2011.2012

3S=2011.2012.2013

S=2011.2012.2013:3

S=2714954572

S=2714954572

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 99.100.101

=> 3S = 99.100.101

=> S = \(\frac{99.100.101}{3}=333300\)

ta xét

\(S\left(n\right)=1.2+2.3+..+n\left(n-1\right)\)

\(\Rightarrow3S\left(n\right)=1.2.3+2.3.3+..+3.n.\left(n-1\right)\)

\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+..+n\left(n-1\right)\left(n+1-\left(n-2\right)\right)\)

\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+n\left(n-1\right)\left(n+1\right)-n\left(n-1\right)\left(n-2\right)\)

\(\Leftrightarrow3S\left(n\right)=n\left(n-1\right)\left(n+1\right)\Rightarrow S\left(n\right)=\frac{n\left(n-1\right)\left(n+1\right)}{3}\)

Áp dụng ta có \(S\left(100\right)=\frac{99.100.101}{3}=333300\)